facas9

2021-01-17

Guided Proof Let ${v}_{1},{v}_{2},....{V}_{n}$ be a basis for a vector space V.
Prove that if a linear transformation $T:V\to V$ satisfies
then T is the zero transformation.
To prove that T is the zero transformation, you need to show that $T\left(v\right)=0$ for every vector v in V.
(i) Let v be the arbitrary vector in V such that $v={c}_{1}{v}_{1}+{c}_{2}{v}_{2}+\cdots +{c}_{n}{V}_{n}$
(ii) Use the definition and properties of linear transformations to rewrite T(v) as a linear combination of $T\left({v}_{j}\right)$ .
(iii) Use the fact that $T\left({v}_{j}\right)=0$
to conclude that $T\left(v\right)=0,$ making T the zero transformation.

sweererlirumeX

Expert

a)Given:
The linear transformation $T:V\to V$
represented as
Approach:
Consider an arbitrary $v={v}_{1},{v}_{2},...,{v}_{n}$ is basis for V.
The function T is said to be linear transformation if it satisfies the vector addition and scalar multiplication properties.
The linear transformation is given by,
$T\left({v}_{i}\right)=0,i=1,2,\cdots ,n...\left(1\right)$
Calculation:
As the vector set v is the subspace of V, the vector v can be written linear combination.
Write the subspace vas linear combination.
$v={c}_{1}{v}_{1}+{c}_{2}{v}_{2}+\cdots +{c}_{n}{v}_{n}\cdots ,\left(2\right)$
Here, ${c}_{1},{c}_{2},\cdots {c}_{n}$ are arbitrary scalars.
Conclusion:
Hence, it is proved above that the set $\left({v}_{1},{v}_{2},\cdots {v}_{n}\right)$ is represented as
$v={c}_{1}{n}_{1}+{c}_{2}{v}_{2},+\cdots +{c}_{n}{v}_{n}.$
b)Given:
The linear transformation $T:V\to V$
represented as
Approach:
Consider an arbitrary $v={v}_{1},{v}_{2},\cdots ,{v}_{n}$ is basis for V.
The function T is said to be linear transformation if it satisfies the vector addition and scalar multiplication properties.
The linear transformation is given by,
$T\left({v}_{i}\right)=0,i=1,2,\cdots ,n\cdots \left(1\right)$
The vector additinon is given by,
$T\left(u+v\right)=T\left(u\right)+T\left(v\right)$
The scalar multiplication is given by,
$T\left(cu\right)=cT\left(u\right)$
Calculation:
As the vector set v is the subspace of V, the vector v can be written linear combination.
Write the subspace vas linear combination.
$T\left(v\right)=\left({c}_{1}{v}_{1}+{c}_{2}{v}_{2}+\cdots +{c}_{n}{v}_{n}\right)$
$=T\left({c}_{1}{v}_{1}\right)+T\left({c}_{2}{v}_{2}\right)+\cdots +T\left({c}_{n}{v}_{n}\right)$
$={c}_{1}T\left({v}_{1}\right)+{c}_{2}T\left({v}_{2}\right)+\cdots +{c}_{n}T\left({v}_{n}\right)....\left(3\right)$
Conclusion:
The transformation form of linear combination $v={c}_{1}{v}_{1}+{c}_{2}{v}_{2}+\cdots +{c}_{n}{v}_{n}$ is
$T\left(v\right)={c}_{1}T\left({v}_{1}\right)+{c}_{2}T\left({v}_{2}\right)+\cdots +{c}_{n}T\left({v}_{n}\right)$
c) Given:
The linear transformation $T:V\to V$
represented as
Approach:
Consider an arbitrary $v={v}_{1},{v}_{2},\cdots ,{v}_{n}$ is basis for V.
The function T is said to be linear transformation if it satisfies the vector addition and scalar multiplication properties.
The linear transformation is given by,
$T\left({v}_{i}\right)=0,i=1,2,\cdots ,n\cdots \left(1\right)$
The vector additinon is given by,
$T\left(u+v\right)=T\left(u\right)+T\left(v\right)$
The scalar multiplication is given by,
$T\left(cu\right)=cT\left(u\right)$
Calculation:
Solve formula (3) with use of formula(1)
$T\left(v\right)={c}_{1}T\left({v}_{1}\right)+{c}_{2}T\left({v}_{2}\right)+\cdots +{c}_{n}T\left({v}_{n}\right)$
$={c}_{1}\left(0\right)+{c}_{2}\left(0\right)+\cdots +{c}_{n}\left(0\right)$
= 0
From above calculation is is clear linear transformation $T:V\to V$
satisfies

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